(1) To prove: $\angle A = 90^\circ$.

Starting with the given equation,

$$a(\cos B + \cos C) = b + c,$$

and applying the cosine rule, we get

$$a \left( \frac{a^2 + c^2 - b^2}{2ac} + \frac{a^2 + b^2 - c^2}{2ab} \right) = b + c.$$

After simplification, this leads to:

$$(b + c)(a^2 - b^2 - c^2) = 0.$$

Since $b + c > 0$, it implies that

$$a^2 = b^2 + c^2.$$

Therefore, by the converse of the Pythagorean theorem, we have $\angle A = 90^\circ$.

$$\boxed{ \angle A = 90^\circ }$$

(2) Given that the radius of the circumcircle of $\triangle ABC$ is 1 and $\angle A = 90^\circ$, according to the definition of the sine function for a right triangle inscribed in a circle with radius 1, we get $a = 2$, because the hypotenuse corresponds to the diameter of the circumcircle.

Now we find $b + c$ as follows:

$$b + c = 2(\sin B + \cos B) = 2\sin(B + 45^\circ).$$

Since $0 < B < 90^\circ$, it follows that $45^\circ < B + 45^\circ < 135^\circ$. Therefore, $\sin(B + 45^\circ)$ increases on this interval, which gives us
$$2 < b + c \leq 2\sqrt{2}.$$

Adding $a = 2$ into the inequality, the range for the perimeter $p = a + b + c$ is

$$4 < p = a + b + c \leq 2 + 2\sqrt{2}.$$

Thus, the range for the perimeter of $\triangle ABC$ is $(4, 2 + 2\sqrt{2}]$.

$$\boxed{ \text{Perimeter range of } \triangle ABC \text{ is } (4, 2 + 2\sqrt{2}] }$$